Question

In: Statistics and Probability

Snow avalanches can be a real problem for travelers in the western United States and Canada....

Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Suppose slab avalanches studied in a region of Canada had an average thickness of μ = 68 cm. The ski patrol at Vail, Colorado, is studying slab avalanches in its region. A random sample of avalanches in spring gave the following thicknesses (in cm).

59 51 76 38 65 54 49 62
68 55 64 67 63 74 65 79

(i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)

x = cm
s = cm


(ii) Assume the slab thickness has an approximately normal distribution. Use a 1% level of significance to test the claim that the mean slab thickness in the Vail region is different from that in the region of Canada.

(a) What is the level of significance?


State the null and alternate hypotheses.

H0: μ < 68; H1: μ = 68 H0: μ ≠ 68; H1: μ = 68     H0: μ = 68; H1: μ > 68 H0: μ = 68; H1: μ < 68 H0: μ = 68; H1: μ ≠ 68


(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The standard normal, since we assume that x has a normal distribution and σ is known. The standard normal, since we assume that x has a normal distribution and σ is unknown.     The Student's t, since we assume that x has a normal distribution and σ is unknown. The Student's t, since we assume that x has a normal distribution and σ is known.


What is the value of the sample test statistic? (Round your answer to three decimal places.)

Solutions

Expert Solution

I HOPE ITS HELPFUL TO YOU IF YOU HAVE ANY DOUBTS PLS COMMENTS BELOW..I WILL BE THERE TO HELP YOU ...ALL THE BEST

AS FOR GIVEN DATA..

i)

x

(x-)^2

59

7.910156

51

116.9102

76

201.2852

38

567.0352

65

10.16016

54

61.03516

49

164.1602

62

0.035156

68

38.28516

55

46.41016

64

4.785156

67

26.91016

63

1.410156

74

148.5352

65

10.16016

79

295.4102

Total

989

1700.438

ii)

a) level of significance ==0.05

b)

n = 16

= 61.81

s = 10.65

Null and alternative hypothesis is

H0 : u = 67

H1 : u ≠ 67

Here population standard deviation is not known so we use t-test statistic.

Test statistic is

c)

Degrees of freedom = n - 1 = 16 - 1 = 15 , α=0.05

P-value = 0.0702        ( using t table)

P-value > α , Fail to Reject H0

I HOPE YOU UNDERSTAND..PLS RATE THUMBS UP ITS HELPS ME ALOT..

THANK YOU...!!


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